Let $f(x)=\cot(e^{ 2x})$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2\csc^2(e^{2x})e^{2x}$ (Choice B) B $\csc^2(e^{2x})e^{2x}$ (Choice C) C $2x\cot(e^{2x})e^x$ (Choice D) D $4x\cdot e^{2x}\csc(e^x)$
Answer: $f$ is a composition of three functions! Let... $u(x)=\cot(x)$ $v(x)=e^{ x}$ $w(x)=2x$... then $f(x)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $f'(x)$, we will need to use the chain rule twice! $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=-\csc^2(x)$ $v'(x)=e^{ x}$ $w'(x)=2$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={-\csc^2(e^{2x})}\cdot{ e^{ 2x}}\cdot{2} \\\\ &=-2\csc^2(e^{2x})e^{2x} \end{aligned}$ In conclusion, $f'(x)=-2\csc^2(e^{2x})e^{2x}$.